Answer:
T = 1.132 x 10⁴ s
Explanation:
given,
Radius of the orbit = 2.8 times radius of Jupiter
Radius of Jupiter, R_J = 71,490 km
gravitational field at the surface of Jupiter, g_J = 22 N/kg
using Kepler's third law equation
[tex]T^2=\dfrac{4\pi^2}{GM}\ r^3[/tex]
rearranging the equation,
[tex]T^2=\dfrac{4\pi^2}{\dfrac{GM}{r^2}}\ r[/tex]
we know,
[tex]g_J = \dfrac{GM}{r^2}[/tex]
now,
[tex]T^2=\dfrac{4\pi^2}{g_J}\times r[/tex]
[tex]T^2=\dfrac{4\pi^2}{22}\times71490\times 10^3[/tex]
T² = 1.2829 x 10⁸
T = 1.132 x 10⁴ s
The period of the spacecraft's orbit is equal to 1.132 x 10⁴ s
